On Wed, Jan 21, 2009 at 6:58 PM, jimdare wrote:
>
> Dear R-Users,
>
> I created the xyplot below using 10 groups (9 groups + 'Total' of all
> groups) with lty=1:10. I need the 'Total' to be a bold solid line (lty=1)
> where as the 9 groups just need to be distinguishable from each other. As
> yo
Hello,
I have the following model:
lm.7 <- lm(Y ~ F + C1 + C2 , data = EM4)
F is a 4-level factor, the rest are covariates centered at their mean (Y
is a two-column matrix).
I have tried to find functions to give the model-adjusted means
(adjusted at the covariates'means) and their standard dev
Hi,
I looked for you on Reunion.com, but you weren't there. Please
connect with me so we can keep in touch.
-Abu
Do You Know Abu?
YES - Connect with Abu, and see who's searching for you
http://smtp26.mail.reunion.co
Hi,
I looked for you on Reunion.com, but you weren't there. Please
connect with me so we can keep in touch.
-Abu
Do You Know Abu?
YES - Connect with Abu, and see who's searching for you
http://smtp26.mail.reunion.co
Hello. I'm trying to make k-step-ahead forecasts using ARFIMA(0, d, 0)
models by taking the first T+k-1 coefficients in the binomial expansion of
(1-B)^d, regarding (1-B)^d x(T+k) as an AR(T+k-1) on x(T+k), where x(T)
is the series value at time T and k = 1, 2, 3,
. That is, I forecast the
ser
On Tue, Jan 20, 2009 at 11:10 PM, Christopher W. Ryan
wrote:
> I start with a dataframe called xrays. It contains scores on films from
> each of two radiologists. It is in "long" format. I used the reshape
> package to melt a data frame and then cast it into "wide" format, one
> line for each pati
I will really appreciate help with plotting some forest plots. I am trying to
show disparities in hazard ratios (and confidence intervals) for a set of
covariates for 2 treatments (A and B say).
I will to be able to keep the column of text (covariate names) but plot the 2
forest plots on an ex
On CRAN there are a number of packages concerned with fitting extreme
value distributions. More than one refers to the book by Stuart Coles,
which I have been using.
Could anyone with experience in this area offer opinions on the different
packages?
David Scott
__
I read the posting about opening .R files in windows with the Rgui and was
wondering if anything has changed for R-2.8.1 and Windows Vista? I have not
yet been able to get the files to open in the Rgui, the association works
enough to open R but that is about it. Thanks for the input.
-Pete
The dataset for that graphic cannot be that big. Why not use dput() on
it and also include the code?
--
David Winsemius
On Jan 21, 2009, at 9:23 PM, jimdare wrote:
Dear R-Users,
I created the xyplot below using 10 groups (9 groups + 'Total' of all
groups) with lty=1:10. I need the 'Total'
Dear R-Users,
I created the xyplot below using 10 groups (9 groups + 'Total' of all
groups) with lty=1:10. I need the 'Total' to be a bold solid line (lty=1)
where as the 9 groups just need to be distinguishable from each other. As
you can probably see, when the group reaches CRA6 the lty start
Dear R-Users,
I created the xyplot below using 10 groups (9 groups + 'Total' of all
groups) with lty=1:10. I need the 'Total' to be a bold solid line (lty=1)
where as the 9 groups just need to be distinguishable from each other. As
you can probably see, when the group reaches CRA6 the lty start
Dear R-Users,
I created the xyplot below using 10 groups (9 groups + 'Total' of all
groups) with lty=1:10. I need the 'Total' to be a bold solid line (lty=1)
where as the 9 groups just need to be distinguishable from each other. As
you can probably see, when the group reaches CRA6 the lty start
Dear R-Users,
I created the xyplot below using 10 groups (9 groups + 'Total' of all
groups) with lty=1:10. I need the 'Total' to be a bold solid line (lty=1)
where as the 9 groups just need to be distinguishable from each other. As
you can probably see, when the group reaches CRA6 the lty start
Dear R-Users,
I created the xyplot below using 10 groups (9 groups + 'Total' of all
groups) with lty=1:10. I need the 'Total' to be a bold solid line (lty=1)
where as the 9 groups just need to be distinguishable from each other. As
you can probably see, when the group reaches CRA6 the lty start
Dear R-Users,
I created the xyplot below using 10 groups (9 groups + 'Total' of all
groups) with lty=1:10. I need the 'Total' to be a bold solid line (lty=1)
where as the 9 groups just need to be distinguishable from each other. As
you can probably see, when the group reaches CRA6 the lty start
Looking at ?points I see information that appears directed at this
question:
Value pch="." (equivalently pch = 46) is handled specially. It is a
rectangle of side 0.01 inch (scaled by cex). In addition, if cex = 1
(the default), each side is at least one pixel (1/72 inch on the pdf,
posts
Dear R-Users,
I created the xyplot below using 10 groups (9 groups + 'Total' of all
groups) with lty=1:10. I need the 'Total' to be a bold solid line (lty=1)
where as the 9 groups just need to be distinguishable from each other. As
you can probably see, when the group reaches CRA6 the lty start
Try this:
myassign <- function(x, val, env = parent.frame())
assign(deparse(substitute(x)), val, env)
myassign(x, 3)
x # 3
On Wed, Jan 21, 2009 at 8:03 PM, Yi Zhang wrote:
> On Wed, Jan 21, 2009 at 5:57 PM, Gabor Grothendieck
> wrote:
>> Try this:
>>
>> f <- function(env = parent.frame()) env$x
I am using the following script to make .jpg files.
jpeg('plotx.jpg')
ddat <-read.table("file",header=T)
attach(ddat)
tdat<-read.table("file1")
plot(xx1,yy1,type='p',pch=1,col="blue",cex=0.2,xlim=c(0,3.5),ylim=c(-75,75))
points(tdat,col="green",pch=1,cex=0.2)
dev.off()
The problem is that I w
One of:
?by
?aggregate
?ave
Next time include the package name where these functions come from AND
code that creates an example data situation and you will increase your
probability of getting a more prompt and complete reply.
--
David Winsemius
On Jan 21, 2009, at 4:54 PM, Ferry wrote:
Try a search on
"cross correlation" "time series"
HTH,
Jim Porzak
TGN.com
San Francisco, CA
http://www.linkedin.com/in/jimporzak
use R! Group SF: http://ia.meetup.com/67/
On Wed, Jan 21, 2009 at 5:17 PM, Michael wrote:
> Hi all,
>
> Is there a way to study the lead and lag relation of two time
try
install.packages('car')
?car::linear.hypothesis
hth,
Kingsford Jones
On Wed, Jan 21, 2009 at 4:20 PM, Roberto Patuelli
wrote:
> Dear All,
>
> I was wondering if it is possible to generate a regression summary (it does
> not matter at this stage if from an lm or for example a glm estimate)
Hi all,
Is there a way to study the lead and lag relation of two time series?
Let's say I have two time series, At and Bt. Is there a systematic way
of concluding whether it's A leading B or B leading A and by how much?
Thanks!
__
R-help@r-project.org
On Wed, Jan 21, 2009 at 5:57 PM, Gabor Grothendieck
wrote:
> Try this:
>
> f <- function(env = parent.frame()) env$x
Thanks. What if the x in "env$x" is an argument passed in? e.g. f <-
function(x, env=parent.frame()) { #assign to env$x ? }
> g <- function(x=1) f()
> x <- 2
> g() # 1
>
> On Wed, J
Greetings all,
Sorry for posting what is primarily not an R question
(though it will have an R target in due course).
Let F(x) be the CDF of the Normal -- i.e. pnorm(x)
Let f(x) be the density function -- i.e. dnorm(x)
Define G(psi) = Integral[-inf,inf] F(x)*f(x)*exp(x*psi) dx
Is G(psi) a know
Thanks, that's great!
Am 22.01.2009 um 01:18 schrieb Remko Duursma:
I use this function (a lot):
ablinepiece <- function(a=NULL,b=NULL,reg=NULL,from,to,...){
# Borrowed from abline
if (!is.null(reg)) a <- reg
if (!is.null(a) && is.list(a)) {
temp <- as.vector(coeffici
I use this function (a lot):
ablinepiece <- function(a=NULL,b=NULL,reg=NULL,from,to,...){
# Borrowed from abline
if (!is.null(reg)) a <- reg
if (!is.null(a) && is.list(a)) {
temp <- as.vector(coefficients(a))
if (length(temp) == 1) {
a <- 0
Hi,
is there a way to define, that a line drawn via abline() should only
go from for example -2 to 1 on the x-axis (with something working
similiar to xlim()) ?
thanks for any help!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailma
Michael Friendly wrote:
In ?title I see the
plot(cars, main = "")
title(main = list("Stopping Distance versus Speed", cex=1.5, col="red",
font=3))
I can't seem to generalize this to use several colors in a single title.
What I'd like is
in latex-ish
\red{Hair color} \black{ and } \blue{Ey
2009/1/21 Michael Friendly :
> In ?title I see the
>
> plot(cars, main = "")
> title(main = list("Stopping Distance versus Speed", cex=1.5, col="red",
> font=3))
>
> I can't seem to generalize this to use several colors in a single title.
Solution from http://tolstoy.newcastle.edu.au/R/e2/help/07
Dear All,
I was wondering if it is possible to generate a regression summary (it does
not matter at this stage if from an lm or for example a glm estimate) in
which to obtain the joint significance of a set of regressors?
Examples could be looking at the joint significance level of a polynomial
In ?title I see the
plot(cars, main = "")
title(main = list("Stopping Distance versus Speed", cex=1.5, col="red",
font=3))
I can't seem to generalize this to use several colors in a single title.
What I'd like is
in latex-ish
\red{Hair color} \black{ and } \blue{Eye color}
to serve also as
Following the recent NYT article about R, I thought this group is not
only ready for R but ready to take it one step further.
Got models in R? Deploy and score them in ADAPA in minutes on the
Amazon EC2 cloud computing infrastructure!
Zementis ( http://www.zementis.com ) has been working with the
Try this:
f <- function(env = parent.frame()) env$x
g <- function(x=1) f()
x <- 2
g() # 1
On Wed, Jan 21, 2009 at 5:45 PM, Yi Zhang wrote:
> Hello,
>
> I'm writing a function like this:
>
> f<-function(x,y,...) {
> ...
> assign(x,y,envir=?)
> }
>
> I need the caller (of f) 's environment for the
de Jong, S. (1993) SIMPLS: an alternative approach to partial least squares
regression. Chemometrics and Intelligent Laboratory Systems, 18, 251263
Thanks
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read
Hello,
I'm writing a function like this:
f<-function(x,y,...) {
...
assign(x,y,envir=?)
}
I need the caller (of f) 's environment for the "?" so that the
assignment is done at the right place. To be specific, when the code
"f(x,1)" appears in environment A, I need the assignment of 1 to x
happen
Hello,
i'm tring to use a cox's model for a survival analysis. I have a dataset,
this is a part:
VOD SESSO fonte_sct donor RT_CGY STATOBMT TEMPO morto
1 0 F midrelated 1200
CP651
2 0 M mid
Hello R Users,
Suppose I have data with the structure below:
Group_Name Pre_Test Post_Test
Grp_A xxx xxx
Grp_A xxx xxx
Grp_A xxx xxx
...
Grp_B xxx xxx
Grp_B xxx xxx
...
Grp_Z x
My bad. I used system("test -r blah") to see if a file was readable,
forgetting that not all Windows installations have the "test" program (from
Cygwin) installed. I've changed this to use the R function file.access() in
version 2.5, which I've just submitted to CRAN. The Windows binary should b
I am working with a list of dates and I would like to replace each date with
the one that comes after, ie. 1/1/07 will become 1/5/07, 1/5/07 will become
1/7/07, etc. The number of days between my dates always varies, so I can't
just increase each one by 5 days or so. Does anyone know of a way I ca
Dear All,
I have interval data (for Mon-Sun, 00-24h) of an activity and would like
to visually plot them in a matrix-like plot, where color A would be
assigned to the activity, and color X to unspecified time usage. Note
that the activities are not in standardised units (hours or so), but
from
Hi,
I have some code with a bunch of apply/sapply/lapply calls to
different functions. I am trying to profile it using Rprof, however
the resulting summary looks like this (output from 'R CMD Rprof'):
% total % self
totalseconds selfsecondsname
100.00 14.5
Dear Gonçalo,
If DF is your data set, something like this should do the job:
which(DF==yourdate,arr.ind=TRUE)
> # Example
> Dates<-c('01/18/2009','01/19/2009','01/20/2009','01/21/2009')
> DF<-data.frame(
+Site=sample(4),
+ "Site#"=sample(4),
+ Season=sample(4),
+Da
Hi,
I have a data.frame SAMPLES with columns:
Site Site# Season Day1 Day2 Day3
Day1, Day2, Day3 are class "Date", the other columns are numeric or
factor.
I have a date "mydate" that may or may not be listed in my data.frame
and I need to find that out.
If "mydate" is there,
Are you sure that you want to loop over variables for each subset, and do
this within each subset? A simple way to generate summary statistics
(across variables and within categories) is the summaryBy function in the
doBy package:
d=data.frame(wt=rnorm(100,100,10),ht=rnorm(100,2,1),sex=gl(2,1,10
Hi,
This is just for print out so it looks nice. what I have is a loop that
loops over my variables and calculates the mean and the sd for these
variables. Then I need to rbind them so I can stack them up in one file. the
problem is that only the fist header appears and the rest do not. this is
the heatmapCol function of package MKmisc might help ... try:
library(MKmisc)
example("heatmapCol")
Best
Matthias
Liu, Hao [CNTUS] wrote:
Dear All:
I tried to use heatmap.2 to generate hierarchical clustering using the
following command:
heatmap.2(datamatrix, scale="row", trace="none", col=g
It works like a charm,thank you all for your help.
--- On Wed, 1/21/09, jim holtman wrote:
> From: jim holtman
> Subject: Re: [R] seq()
> To: mazatlanmex...@yahoo.com
> Cc: r-h...@stat.math.ethz.ch
> Date: Wednesday, January 21, 2009, 9:40 AM
> Is this what you want:
>
> > x <- seq(1, 52, 2
Hello,
i'm tring to use a cox's model for a survival analysis. I have a dataset,
this is a part:
VOD SESSO fonte_sct donor RT_CGY STATOBMT TEMPO morto
1 0 F midrelated 1200
CP651
2 0 M mid
Hi,
I'm trying to use metaMDS with a dissimilarity matrix of angles, not
Bray-Curtis, and I wanted to know if there is an in-built function to
produce a plot of stress values against dimensions, that could be used to
determine the 'true' dimension of the solution. The number of objects is
only a l
On Wed, 2009-01-21 at 09:29 -0800, Felipe Carrillo wrote:
> HI:
> Could someone help me with the seq function? I have a range of values
> starting from 1 to 52 but I want seq to start at 27 by=2, but when it reaches
> 51 start with with number 1 to 25. is this possible. I can do the basics of
>
Hi Felipe,
concatenate two sequences using c():
c(seq(from=27,to=51,by=2),seq(from=1,to=25,by=2))
HTH,
Stephan
Felipe Carrillo schrieb:
HI:
Could someone help me with the seq function? I have a range of values starting
from 1 to 52 but I want seq to start at 27 by=2, but when it reaches 51
Is this what you want:
> x <- seq(1, 52, 2)
> x
[1] 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43
45 47 49 51
> (x + 26) %% 52
[1] 27 29 31 33 35 37 39 41 43 45 47 49 51 1 3 5 7 9 11 13 15 17
19 21 23 25
>
On Wed, Jan 21, 2009 at 12:29 PM, Felipe Carrillo
wrote:
> HI:
G'day Felipe,
On Wed, 21 Jan 2009 09:29:03 -0800 (PST)
Felipe Carrillo wrote:
> Could someone help me with the seq function? I have a range of values
> starting from 1 to 52 but I want seq to start at 27 by=2, but when it
> reaches 51 start with with number 1 to 25. is this possible. I can do
>
HI:
Could someone help me with the seq function? I have a range of values starting
from 1 to 52 but I want seq to start at 27 by=2, but when it reaches 51 start
with with number 1 to 25. is this possible. I can do the basics of seq() but I
can't figure how to do this one. This is how I want my s
Gregor,
Section 6.1 of the FAQ provides further examples. It does depend on
the editor you are using. Several editors come with add-ons that
support R.
Recently, on the Mac specific r-sig-mac mailing list ( r-sig-...@stat.math.ethz.ch
), Smultron came up.
Regards,
Rob
On Jan 21, 2009, a
So far we have dput(a). Does that represent what you have?
what you want? Can you provide both and a description in words.
See the last line to every message to r-help.
On Wed, Jan 21, 2009 at 11:13 AM, Sergey Goriatchev wrote:
> Dear Dr. Grothendieck,
>
> My purpose in this case it to have the
?try
On Wed, Jan 21, 2009 at 10:55 AM, Brigid Mooney wrote:
> Hi All,
>
> I have an script in R which accepts user inputs for certain parameters,
> particularly dates, which the user inputs as character strings.
> eg:
> > date1 <- "2009-01-21"
>
> The script later parses the input via the as.Dat
What do you want to do with it? Is this just for printing out? What
other types of transformations are you intending to do? Why not just
put them in a 'list' and then write your own specialized print
routine.
On Wed, Jan 21, 2009 at 10:30 AM, SNN wrote:
>
> Hi,
>
> I need to rbind two data fram
Hi,
I need to rbind two data frames. Each one has a header . after the rbind I
would like to keep the header for each and have the two data frames
separated by a line. Is this possible to do in R?
For example
weight_meanweight_sd.dev
> F 14.3 4.932883
> M 34.7 10.692677
>
>
Dear list members,
I am using a cross validation of a generalised linear model (glm). The cv.glm
function (from boot package) returns an error as so-called „delta“ value. I
would like to get to a (cross-validated) squared q, because I want to directly
compare it to the squared correlation coeff
> I do have a data set with some missing values that appear as blanks. I
want
> to fill these blanks with an NA. How can this be done? Thanks for your
help
Please help us to help you. What form are the data in? Are they in a
text file, or are they in R already? What do you mean by 'blanks'?
Dear Dr. Grothendieck,
My purpose in this case it to have the structure of the object on
non-Bloomberg machine the same as that of the object on the Bloomberg
machine.
That way I am sure that whatever code I write away from Bloomberg
machine will work on it when I copy the code to it.
Also, I am v
Is your purpose to change the times in some way?
If z is a zoo series you can change the times like this:
library(zoo)
library(chron)
time(z) <- ...whatever...
e.g.
> z <- zoo(1:3, 11:13)
> z
11 12 13
1 2 3
> time(z) <- as.Date(11:13)
> z
1970-01-12 1970-01-13 1970-01-14
1
Hi All,
I have an script in R which accepts user inputs for certain parameters,
particularly dates, which the user inputs as character strings.
eg:
> date1 <- "2009-01-21"
The script later parses the input via the as.Date function:
> as.Date(date1)
However, as.Date encounters an error when the
Dear list,
I'm posting the solution to my problem in case others may find this useful.
This code was sent to me by Phil Spector. With a bit of cleaning, it can easily
be converted to a usable format. Thanks to Gabor Grothendieck, David winsemius
and Martin Morgan for also sending possible sol
Dear Dr. Grothendieck,
First of all, I realized I did not load zoo package before I tried the
first str(bldata). If I load zoo and then do str(bldata) I get the
following:
'zoo' series from 7305 to 14609
Data: num [1:5219, 1:12] 91.9 91.8 91.7 91.8 91.7 ...
- attr(*, "dimnames")=List of 2
..
Try this also:
is.na(my.data$var) <- which(my.data$var == "")
On Wed, Jan 21, 2009 at 12:33 PM, kayj wrote:
>
> Hi,
>
> I do have a data set with some missing values that appear as blanks. I want
> to fill these blanks with an NA. How can this be done? Thanks for your
> help
>
> --
> View this
In addition to this, is it in anyway possible to get the variance in the
prediction graph as a shade similar to the GAM plots?
2009/1/21 Robbert Langenberg
> Thanks a lot for all the mails!
>
> With all the different examples I now have got it working. The problem
> finally after I had worked ou
Hello everybody!
I have a problem when I try to perform a forecast of an ARIMA model
produced by an auto.arima function. Here is what I'm doing:
c<-auto.arima(fil[[1]],start.p=0,start.q=0,start.P=0,start.Q=0,stepwise=TRUE,stationary=FALSE,trace=TRUE)
# fil[[1]] is time series of monthly data
Dear R-users,
Has anyone written a function for multifractal detrended
fluctuation analysis? The "fractal" package does mono-fractal DFA,
but not multifractal as far as I can tell. The MF-DFA approach is
presented in:
J. W. Kantelhardt, S. Zschiegner, E. Koscielny-Bunde, S. Havlin, A.
B
Hi,
kayj wrote:
> Hi,
>
> I do have a data set with some missing values that appear as blanks. I want
> to fill these blanks with an NA. How can this be done? Thanks for your help
Something like this?
> my.data<-data.frame("var"=c(1,2,5,"",66,4,3,"",67,5,3,2,1,4,32,56,23),
stringsAsFactors=F)
Hello,
I've a time series, T, for an irregular sequence of days (data for bank
holidays weekends have been removed). I would like to aggregate this series
by into weeks using
aggregate(x, by="weeks", FUN=mean).
To do this, I think that x needs to be a timeSeries object, and when I try
to call
Hi,
I do have a data set with some missing values that appear as blanks. I want
to fill these blanks with an NA. How can this be done? Thanks for your help
--
View this message in context:
http://www.nabble.com/filling-blanks-with-NA-tp21584278p21584278.html
Sent from the R help mailing list
Hello everybody!
We have a problem with a linux server that crashes when we try to read large
datasets in R.
The R code is as followed:
komplett <- read.spss("komplett2003aar.sav", to.data.frame =TRUE, reencode
="Latin1")
The information about the linux server is:
Linux version 2.6.24-19-gene
Thanks a lot for all the mails!
With all the different examples I now have got it working. The problem
finally after I had worked out the newd2 as supplied by Gavin was that I
found the plot so strange looking. I have altered my newd2 a bit, and after
your e-mails Simon I figured out how to finall
Another possibility is Reduce:
Reduce(function(x,y,by='pos')merge(x,y,by='pos'),mylist)
pos data.x data.y data
1 A 2 69
2 B 6 23
3 C 3 96
4 D 1 72
5 E 9 51
- Phil Spector
Please reduce your examples down to small amounts of data and use dput so
that they are reproducible.
On Wed, Jan 21, 2009 at 9:32 AM, Sergey Goriatchev wrote:
> Dear all,
>
> I have a zoo object that has following structure:
>
>> str(bldata)
> zoo [1:5219, 1:12] 91.9 91.8 91.7 91.8 91.7 ...
>
Try this:
> x <- list(NULL, 1:10, 25, NULL, "more")
> x
[[1]]
NULL
[[2]]
[1] 1 2 3 4 5 6 7 8 9 10
[[3]]
[1] 25
[[4]]
NULL
[[5]]
[1] "more"
> which(sapply(x, is.null))
[1] 1 4
>
On Wed, Jan 21, 2009 at 7:42 AM, diego Diego wrote:
> Hello everybody!
> I have a list of length 5000
Dear all,
I have a zoo object that has following structure:
> str(bldata)
zoo [1:5219, 1:12] 91.9 91.8 91.7 91.8 91.7 ...
- attr(*, "index")=Classes 'dates', 'times' atomic [1:5219] 7305
7306 7307 7308 7309 ...
.. ..- attr(*, "format")= chr "m/d/y"
.. ..- attr(*, "origin")= Named num [1:3]
Hello everybody!
I have a list of length 5000 whose components are (mostly) "ts" objects,
but some these components are intentionally left empty, ie, they are "NULL"
components of the list. My question is how can I get the position of these
null components in a effective way (I'm trying to avoid a
Dear Juliet,
I found out that the issue of matching more than two vectors can be avoided by
pasting vectors into one using paste(). Also, it is not actually necessary for
my problem to use match(). I was looking for a method to match points by
coordinates, this can be done be using equal coordi
Thank you all for your help!
Sergey
On Wed, Jan 21, 2009 at 13:25, Gabor Grothendieck
wrote:
> This will give you 6 months after today. Use negative
> numbers to move backwards:
>
>> d <- Sys.Date()
>> seq(d, length = 2, by = paste(6, "months"))[2]
> [1] "2009-07-21"
>
>
> On Wed, Jan 21, 2009
merge.zoo can do a multiple merge. We create a list of
zoo objects, mylist.z, and then perform the merge:
> library(zoo)
> mylist.z <- lapply(mylist, function(x) zoo(x$data, as.character(x$pos)))
> do.call(merge.zoo, mylist.z)
df1 df2 df3
A 2 6 9
B 6 2 3
C 3 9 6
D 1 7 2
Hi Antje,
Try this:
merge(merge(mylist[[1]], mylist[[2]], by = "pos"), mylist[[3]])
On Wed, Jan 21, 2009 at 11:19 AM, Antje wrote:
> Henrique Dallazuanna schrieb:
>
>> Try this also:
>>
>> cbind(pos = mylist$df1$pos, data.frame(mylist)[grep("data",
>> names(data.frame(mylist)))])
>>
>
> Hi He
Henrique Dallazuanna schrieb:
Try this also:
cbind(pos = mylist$df1$pos, data.frame(mylist)[grep("data",
names(data.frame(mylist)))])
Hi Henrique,
cool solution - that's seems to be the easiest way!
though I thought there should be some possibiliy of multiple merge
Anyway, this will do it
Have you tried specifying an absolute path in prefix.string instead of a
relative one?
HTH,
Thierry
Good point! This actually solves the problem, I just wanted to avoid
this as I wanted to send it to someone else... so I just reordered the
files and it works.
Solutions given by Duncan
Try this also:
cbind(pos = mylist$df1$pos, data.frame(mylist)[grep("data",
names(data.frame(mylist)))])
On Wed, Jan 21, 2009 at 6:19 AM, Antje wrote:
> Hi there,
>
> I have a list of dataframes (generated by reading multiple files) and all
> dataframes are comparable in dimension and column na
Li, Hua wrote:
Dear R helpers:
Let's say I have some data X,
X <- runif(1000, 1, 100)
pdf('X.pdf', width=100,height=5)
hist(X, breaks=1000)
dev.off()
I find that, on x-axis the coordinates are 0e+00, 2e+09, 4e+09, 6e+09,
8e+09, 1e+10. Only five numbe
This will give you 6 months after today. Use negative
numbers to move backwards:
> d <- Sys.Date()
> seq(d, length = 2, by = paste(6, "months"))[2]
[1] "2009-07-21"
On Wed, Jan 21, 2009 at 6:27 AM, Sergey Goriatchev wrote:
> Dear Gabor,
>
> Thanks for that!
> Still, it is not really similar to
Thomas Lumley wrote:
> On Tue, 20 Jan 2009, Stavros Macrakis wrote:
>
>> I'm rather confused by the semantics of factors.
>>
>
>>
>> It is all very confusing. Of course, most of this behavior is
>> documented and is easily determined by experimentation, but it would
>> be easier to learn and tea
On Wed, 21 Jan 2009, Sergey Goriatchev wrote:
Dear Prof. Ripley,
Thank you for help.
Yes, that is an interesting question you pose. I already thought
myself how February should be handled, as EDATE(31.08.2008; -6)
returns 29.02.2008.
In Excel it is not a problem, since this nonexisting date is
Dear Prof. Ripley,
Thank you for help.
Yes, that is an interesting question you pose. I already thought
myself how February should be handled, as EDATE(31.08.2008; -6)
returns 29.02.2008.
In Excel it is not a problem, since this nonexisting date is then used
in VLOOKUP function where one can have
Team,
I have a time series data with lot of seasonality and trends. I am familiar
with some forecasting techniques, but for this set, I believe STL would be
the best. Please let me know any tips to use STL using R.
Best,
Kishore
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On Wed, 21 Jan 2009, Sergey Goriatchev wrote:
Dear Gabor,
Thanks for that!
Still, it is not really similar to how EDATE works.
With julian(Sys.Date(), Sys.Date() - 10) one moves 10 days back.
The problem is that I need to move by months, not by days, as months
have different number of days.
I
1) Please update your R as requested in the posting guide, and provide
the 'at a mimimum' information requested.
2) Try reading ?postscript and setting the encoding: I suspect yours
is not one of the cases that are guessed correctly.
3) If all else fails, give a 'commented, minimal, self-cont
Dear Gabor,
Thanks for that!
Still, it is not really similar to how EDATE works.
With julian(Sys.Date(), Sys.Date() - 10) one moves 10 days back.
The problem is that I need to move by months, not by days, as months
have different number of days.
I need to come to the same day when I move backward
"enrico.fosco...@libero.it" napsal dne
21.01.2009 11:54:49:
> I can't use the function "outer" because my function fun() doesn't take
vectors
> as arguments.
If you can not
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http:/
Good morning to all,
I should evaluate a function for every cell of a given
matrix with n rows and n columns.
This function, named fun(), has got two
arguments: the number of the row and the number of the column which
characterized every single cell.
So the result should be
fun(1,1) fun(1,2
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